# Thread: Eigenvalues of Linear operator

1. ## Eigenvalues of Linear operator

Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
$d/dt" alt="d/dt" />----------> $L_2[0,1]$ where the domain D is given by D{f belongs to $L_2[0,1]$: f is continuously differentiable and f(0)=0)}
I get $f(x)=e^{\lambda x}$ after solving first order linear differential equation. How do i find $\lambda$?
thanks

2. Originally Posted by charikaar
Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
$d/dt" alt="d/dt" />----------> $L_2[0,1]$ where the domain D is given by D{f belongs to $L_2[0,1]$: f is continuously differentiable and f(0)=0)}
I get $f(x)=e^{\lambda x}$ after solving first order linear differential equation. How do i find $\lambda$?
thanks
our operator is just differentiation so, yes, $\frac{d e^{\lambda x}}{dx}= \lambda e^{\lambda x}$. More generally, if $\frac{dy}{dx}= \lambda y$, then $y= Ce^{\lambda 0}$ $e^{\lambda x}$ is an "eigenvector" and $\lambda$ is an eigenvalue. As for finding $\lambda$, you don't! That equation is true for all $\lambda$- any real number is an eigenvalue.

As for whether or not it is "self adjoint", the definition of self-adjoint is that <Af, g>= <f, Ag> for any vectors f, g. Here, I presume you are using the "standard" inner product, $= \int_0^1 f(x)g(x)dx$ so that we want $\int_0^x \frac{df}{dx} g dt$. Use "integration by parts" with u= g and dv= (df/dx)dx. Then du= (dg/dx)dx and v= f.

$\int u dv= uv- \int vdu= f(x)g(x)- \int f (dg/dx) dx= $ provided f(1)g(1)- f(0)g(0)= 0.

3. I'm curious on one thing. Don't we require that $f(0) = 0$?

$L_2[0,1]$ is the Hilbert space of complex square integrable functions on [0,1] and D={f belongs to $L_{2}[0,1]$ : f is continuously differentiable and f(0)=1)}. so $\lambda$ includes complex numbers also? L is not self-adjoint because there is minus in front of the integral right?